The Force Between Two Nucleons
The Force Between Two Nucleons
The Force Between Two Nucleons
◎ The deuteron
● Nucleon-Nucleon scattering
Fundamental Forces
A few properties of nucleon-nucleon force:
The simplest nucleus in nature is that of the hydrogen isotope, deuterium. Known as
the “deuteron,” the nucleus consists of one proton and one neutron. Due to its
simplicity, the deuteron is an ideal candidate for tests of our basic understanding of
nuclear physics. Recently, scientists have been studying the intrinsic shape of the
deuteron. Dominated by three components describing the interactions of the quark
components of the neutron and proton, its shape is not spherical. Recent tests have
shown no deviations in the predictions of standard nuclear physics.
www.physicscentral.com/.../2002/deuteron.html
The Deuteron - Angular momentum
1. In analogy with the ground state of the hydrogen atom, it is reasonable to assume
that the ground state of the deuteron also has zero orbital angular momentum L = 0
3. The implication is that two nucleons are not bound together if their spins are anti-
parallel, and this explains why there are no proton-proton or neutron-neutron bound
states (more later).
4. The parallel spin state is forbidden by the Pauli exclusion principle in the case of
identical particles
(4)
(r ) = R(r )Ylm ( , )
(6)
Substitute R(r) = u(r)/r in to the Schrödinger’s equation the function u(r) satisfies the
following equation ;
2 d 2u l (l + 1) 2
− + V (r ) + u (r ) = Eu(r )
2m dr 2 2mr 2 (7)
For the case of a three dimensional square well potential with zero angular
momentum (l = 0), which we use as the model potential for studying the ground
state of the deuteron, the Schrödinger’s equation can be simplified into:
2 d 2u
− − V0u (r ) = Eu(r ) , for r < R
2m dr 2 (12)
2 2
d u
− = Eu(r ) , for r > R
2m dr 2
I. When r < R
2 d 2u
The Schrödinger’s equation is − − V0u (r ) = Eu(r ) (13)
2m dr 2
This equation can be rearranged into:
d 2u 2m( E + V0 )
+ u (r ) = 0 k1
2
k1 with (14)
dr 2 2
And the solution is u(r ) = A sin k1r + B cos k1r (15)
u (r )
To keep the wave function finite for r → 0 lim (r ) = lim =0
r →0 r →0 r
2 d 2u
The Schrödinger’s equation is: − = Eu(r ) (17)
2m dr 2
− 2mE
with k2 =
2
u (r ) = Ce − k 2 r (19)
Applying the continuity conditions on u(r) and du/dr at r = R, we obtain
R = 2.1 fm
V0 = 35 MeV
Here we show the deuteron wave
function for R = 2.1 fm. The
exponential joins smoothly to the
sine at r = R, so that both u(r) and
du/dr are continuous.
If the nucleon-nucleon force were just a bit weaker the deuteron bound
state would not exist at all. In this situation the whole universe would be
all quite different from the one we are observing.
Spin and parity of the deuteron
I = sn + sp + l (21)
● Since we know that the parity of the deuteron is even and the parity associated
with orbital motion is determined by (-1)l we are able to rule out some options.
● Orbital angular momentum l = 0 and l = 2 give the correct parity determined from
experimental observations.
● The observed even parity allows us to eliminate the combinations of spins that
include l =1, leaving l = 0 and l = 2 as possibilities.
The magnetic dipole moment of the deuteron
If the l = 0 is perfectly correct description for the deuteron, there should be no orbital
contribution to the magnetic moment. We can assume that the total magnetic moment is
simply the combination of the neutron and proton magnetic moments:
= n + p
g sn N g sp N
= sn + sp (22)
where gsn = -3.826084 and gsp = 5.585691.
If we take the observed magnetic moment to be the z component of μ
1
when the spins have their maximum value (+ )
2
1
= N ( g sn + g sp )
2 (23)
= 0.879804 N
= as 2 (l = 0) + ad 2 (l = 2) (25)
When the mixed wave function [equation (24)] is used to calculate the quadrupole
moment of the deuteron (Q) the calculation gives two contribution terms. One is
proportional to (ad)2 and another proportional to the cross-term (asad).
2 1 2 2
Q= as ad r 2 − ad r (28)
10 sd 20 dd
where r2 = r 2 Rs (r ) Rd (r )r 2 dr r2 = r 2 Rd (r ) Rd (r )r 2 dr
sd dd
To calculate Q we must know the deuteron d-state wave function and it is obtainable from
the realistic phenomenological potentials. The d-state admixture is of several percent in this
calculation and is consistent with the 4% value deduced from the magnetic moment.
Some comments concerning the d-state admixture obtained from
the studies of magnetic moment μ and the quadrupole moment Q:
1. This good agreement between the d-state admixtures deduced from μ and Q should
be regarded as a happy accident and not taken too seriously. In the case of the
magnetic dipole moment, there is no reason to expect that it is correct to use the
free-nucleon magnetic moments in nuclei.
2. Spin-orbit interactions, relativistic effects, and meson exchanges may have greater
effects on μ than the d-state admixture (but may cancel one another’s effect).
3. For the quadrupole moment, the poor knowledge of the d-state wave function makes
the deduced d-state admixture uncertain.
4. Other experiments, particularly scattering experiments using deuterons as targets,
also give d-state admixtures in the range of 4%. Thus our conclusions from the
magnetic dipole and electric quadrupole moments may be valid after all.
2. Along the surface of any spherical scattered wave front, the diffraction is responsible
for the variation in intensity of the radiation. The intensity thus depends on angular
coordinates θ and φ.
3. A radiation detector placed at any point far from the target would record both incident
and scattered waves.
To solve the nucleon-nucleon scattering problem using quantum mechanics we
assume the nuclear interaction by a square-well potential, as we did for the
deuteron.
In fact, the only difference between this calculation and that of the deuteron
is that here we concerned with free incident particles with E > 0.
The minus sign between the two terms keeps the incident wave function
finite for r → 0, and using the coefficient A for both terms sets the
amplitudes of the incoming and outgoing waves to be equal.
We further assume that the scattering does not create or destroy particles, and
thus the amplitudes of the eikr and e-ikr terms should be the same.
All that can result from the scattering is a change in phase of the outgoing wave:
In terms of scattered , the current of scattered particles per unit area can thus be
calculated by the following equation:
*
jscattered = * − (40)
2im r r
The scattered current is uniformly distributed over a sphere of radius r. The probability dσ
that an incident particle is scattered into dΩ is the ratio of the scattered current to the
incident current:
( jscattered)(r 2dΩ) (41.a)
d =
jincident
By subtracting the incident part of the wave function from Eq. (43)
we have the scattered wave.
scattered = − incident
A 2i 0 eikr (44)
= (e − 1)
2ik r
Using Eq. (40) the current of scattered particles per unit area is:
2
* A
jscattered = * − = sin 2 0 (45)
2mi r r mkr
2
2
k A
The incident current is jincident = (46)
m
d sin 2 0
The differential cross section is = (47)
dΩ k2
In general, dσ/ dΩ varies with direction over the surface of the
sphere; in the special case of l = 0 scattering, dσ/ dΩ is
constant and the total cross section σ is
d sin 2 0 4π sin 2 δ0
= dΩ = dΩ = (48)
dΩ k2 k2
Assume that the incident energy is small, say E ≦ 10 keV and take V0
= 35 MeV from our analysis of the deuteron bound state.
If we let the right side of Equation (36) equal –α then = −k1 cot k1R
(50)
cos k2 R + ( / k2 ) sin k2 R
A bit of trigonometric manipulation gives sin 2 0 = (51)
1 + 2 / k2
2
4
and so =
2
cos k R + sin k R
2
(52)
k2 +
2 2
k2
and
Which requires at and as to be not only different but also opposite in sign:
These results compare well with the theoretical estimates of deuteron bound states.
-
PROTON PROTON AND NEUTRON -NEUTRON
INTERACTIONS
total (anti-symmetric) total = space * spin
If we again consider only low-energy scattering, so that l = 0, interchanging
the spatial coordinates of the two particles gives no change in sign i.e. space = Sym
• Short range
• Attractive at long range
• Repulsive at short range
• Spin dependent
• Charge independent
• Spin-orbit force
• Tensor force
• Conserve parity
SHORT RANGE
• The range of the nuclear force must be smaller than the size of the
atom. All atomic and molecular spectra can be explained by the
electromagnetic (em) force alone.
• The binding energy/nucleon is almost constant
Attractive at long range
V (MeV)
Spin dependence
Spin singlet and spin triplet phase shifts are a little different with the spin
triplet being more attractive than the singlet (they both become negative at
large momentum)
2S+1L
J
Q > 0, observed
• A tensor force is a non central force, because it does not just depend on the
relative distance between two particles, but also on the relative orientation
V = V (r) and on its orientation with respect to the spins of the nucleons.
• Tensor force: depends on the angle between relative distance vector and the
spins of the nucleons, in analogy to the force between two magnetic dipoles
The Nucleon -Nucleon Force is Charge Symmetric
This means that the proton-proton interaction is identical to the neutron-neutron
interaction, after we correct for the Coulomb force in the proton-proton system.
ISOSPIN
(This plays a very important role also in determining the correct order of energy
levels in nuclear spectra – more later!)
Spin-orbit interaction
Spin-orbit interaction
Spin-orbit interaction
Most general form for nuclear force
Yukawa theory: One-pion-exchange potential
Yukawa theory: One-pion-exchange potential
T=K.E.
Yukawa theory: One-pion-exchange potential
Yukawa theory: One-pion-exchange potential